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3 years ago

14

Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A. Decide in each case whether R

is an equivalence relation, justifying your answers. In the case that R is an equivalence relation, find the equivalence classes. (a) (a, b)R(c, d) if and only if ad = bc

1 answer:

andre [41]3 years ago

5 0

Answer:

Step-by-step explanation:

Let A = R−{0}, the set of all nonzero real numbers, and consider the following relations on A × A.

Given that (a,b) R (c,d) if $ad=bc$

Or (a,b) R (c,d) if determinant

$\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =0$

a) Reflexive:

We have (a,b) R (a,b) because ab-ab =0 Hence reflexive

b) Symmetric

(a,b) R (c,d) gives ad-bc =0

Or da-cb =0 or cb-da =0 Hence (c,d) R(a,b). Hence symmetric

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2 years ago

In ΔUVW, the measure of ∠W=90°, UV = 6.2 feet, and WU = 1.2 feet. Find the measure of ∠V to the nearest tenth of a degree.

Crazy boy [7]

Answer:

m<V = 11.2 degrees

Step-by-step explanation:

In ΔUVW, the measure of ∠W=90°, UV = 6.2 feet, and WU = 1.2 feet.

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2 years ago

Read 2 more answers

If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

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